Difference between revisions of "1983 AIME Problems/Problem 8"

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== Problem ==
 
== Problem ==
 
What is the largest <math>2</math>-digit [[prime]] factor of the integer <math>n = {200\choose 100}</math>?
 
What is the largest <math>2</math>-digit [[prime]] factor of the integer <math>n = {200\choose 100}</math>?
  
== Solution ==
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==Solution==
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=== Solution 1 ===
  
 
Expanding the [[combination|binomial coefficient]], we get <math>{200 \choose 100}=\frac{200!}{100!100!}</math>. Let the required prime be <math>p</math>; then <math>10 \le p < 100</math>. If <math>p > 50</math>, then the factor of <math>p</math> appears twice in the denominator. Thus, we need <math>p</math> to appear as a factor at least three times in the numerator, so <math>3p<200</math>. The largest such prime is <math>\boxed{061}</math>, which is our answer.
 
Expanding the [[combination|binomial coefficient]], we get <math>{200 \choose 100}=\frac{200!}{100!100!}</math>. Let the required prime be <math>p</math>; then <math>10 \le p < 100</math>. If <math>p > 50</math>, then the factor of <math>p</math> appears twice in the denominator. Thus, we need <math>p</math> to appear as a factor at least three times in the numerator, so <math>3p<200</math>. The largest such prime is <math>\boxed{061}</math>, which is our answer.
  
== Solution 2: Clarification of Solution 1 ==
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=== Solution 2: Clarification of Solution 1 ===
  
 
We know that  
 
We know that  
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So basically, <math>p</math> is the largest prime number such that
 
So basically, <math>p</math> is the largest prime number such that
<cmath>\lfloor\frac{200}{p}\rfloor>3</cmath>
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<cmath>\left \lfloor\frac{200}{p}\right \rfloor>3</cmath>
 
Since <math>p<\frac{200}{3}=66.66...</math>, the largest prime value for <math>p</math> is <math>p=\boxed{61}</math>
 
Since <math>p<\frac{200}{3}=66.66...</math>, the largest prime value for <math>p</math> is <math>p=\boxed{61}</math>
  

Latest revision as of 20:23, 14 January 2023

Problem

What is the largest $2$-digit prime factor of the integer $n = {200\choose 100}$?

Solution

Solution 1

Expanding the binomial coefficient, we get ${200 \choose 100}=\frac{200!}{100!100!}$. Let the required prime be $p$; then $10 \le p < 100$. If $p > 50$, then the factor of $p$ appears twice in the denominator. Thus, we need $p$ to appear as a factor at least three times in the numerator, so $3p<200$. The largest such prime is $\boxed{061}$, which is our answer.

Solution 2: Clarification of Solution 1

We know that \[{200\choose100}=\frac{200!}{100!100!}\] Since $p<100$, there is at least $1$ factor of $p$ in each of the $100!$ in the denominator. Thus there must be at least $3$ factors of $p$ in the numerator $200!$ for $p$ to be a factor of $n=\frac{200!}{100!100!}$. (Note that here we assume the minimum because as $p$ goes larger in value, the number of factors of $p$ in a number decreases,)

So basically, $p$ is the largest prime number such that \[\left \lfloor\frac{200}{p}\right \rfloor>3\] Since $p<\frac{200}{3}=66.66...$, the largest prime value for $p$ is $p=\boxed{61}$

~ Nafer

See Also

1983 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions